与えられた逆三角関数の方程式を解き、$x$の値を求めます。 (1) $\sin^{-1}\frac{4}{5} + \sin^{-1}\frac{5}{13} = \cos^{-1}x$ (2) $\tan^{-1}\frac{1}{2} - \tan^{-1}\frac{3}{4} = \sin^{-1}x$ (3) $\sin^{-1}\frac{2}{\sqrt{13}} + \cos^{-1}\frac{3}{5} = \tan^{-1}x$ (4) $\cos^{-1}\frac{2}{3} + \cos^{-1}\frac{1}{4} = \sin^{-1}x$

はい、承知いたしました。三角関数の加法定理を用いて、逆三角関数の方程式を解く問題ですね。各問題について順に解説します。

1. 問題の内容

与えられた逆三角関数の方程式を解き、$x$の値を求めます。

(1) $\sin^{-1}\frac{4}{5} + \sin^{-1}\frac{5}{13} = \cos^{-1}x$

(2) $\tan^{-1}\frac{1}{2} - \tan^{-1}\frac{3}{4} = \sin^{-1}x$

(3) $\sin^{-1}\frac{2}{\sqrt{13}} + \cos^{-1}\frac{3}{5} = \tan^{-1}x$

(4) $\cos^{-1}\frac{2}{3} + \cos^{-1}\frac{1}{4} = \sin^{-1}x$

2. 解き方の手順

(1) $\sin^{-1}\frac{4}{5} + \sin^{-1}\frac{5}{13} = \cos^{-1}x$

$\alpha = \sin^{-1}\frac{4}{5}, \beta = \sin^{-1}\frac{5}{13}$ とおくと、

$\sin\alpha = \frac{4}{5}, \sin\beta = \frac{5}{13}$

$\cos\alpha = \sqrt{1 - (\frac{4}{5})^2} = \frac{3}{5}, \cos\beta = \sqrt{1 - (\frac{5}{13})^2} = \frac{12}{13}$

$\sin(\alpha+\beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta = \frac{4}{5}\cdot\frac{12}{13} + \frac{3}{5}\cdot\frac{5}{13} = \frac{48+15}{65} = \frac{63}{65}$

$\alpha + \beta = \sin^{-1}\frac{63}{65}$

$\sin^{-1}\frac{63}{65} = \cos^{-1}x$

$\cos(\sin^{-1}\frac{63}{65}) = x$

$\cos(\sin^{-1}\frac{63}{65}) = \sqrt{1 - (\frac{63}{65})^2} = \sqrt{\frac{65^2 - 63^2}{65^2}} = \sqrt{\frac{(65+63)(65-63)}{65^2}} = \sqrt{\frac{128 \cdot 2}{65^2}} = \sqrt{\frac{256}{65^2}} = \frac{16}{65}$

$x = \frac{16}{65}$

(2) $\tan^{-1}\frac{1}{2} - \tan^{-1}\frac{3}{4} = \sin^{-1}x$

$\alpha = \tan^{-1}\frac{1}{2}, \beta = \tan^{-1}\frac{3}{4}$

$\tan(\alpha - \beta) = \frac{\tan\alpha - \tan\beta}{1 + \tan\alpha\tan\beta} = \frac{\frac{1}{2} - \frac{3}{4}}{1 + \frac{1}{2}\cdot\frac{3}{4}} = \frac{\frac{2-3}{4}}{1 + \frac{3}{8}} = \frac{-\frac{1}{4}}{\frac{11}{8}} = -\frac{1}{4}\cdot\frac{8}{11} = -\frac{2}{11}$

$\tan^{-1}(-\frac{2}{11}) = \sin^{-1}x$

$\sin(\tan^{-1}(-\frac{2}{11})) = x$

$\tan\theta = -\frac{2}{11}$. $\sin\theta = \frac{\tan\theta}{\sqrt{1+\tan^2\theta}} = \frac{-\frac{2}{11}}{\sqrt{1+(\frac{2}{11})^2}} = \frac{-\frac{2}{11}}{\sqrt{\frac{121+4}{121}}} = \frac{-\frac{2}{11}}{\frac{\sqrt{125}}{11}} = -\frac{2}{\sqrt{125}} = -\frac{2}{5\sqrt{5}} = -\frac{2\sqrt{5}}{25}$

$x = -\frac{2\sqrt{5}}{25}$

(3) $\sin^{-1}\frac{2}{\sqrt{13}} + \cos^{-1}\frac{3}{5} = \tan^{-1}x$

$\alpha = \sin^{-1}\frac{2}{\sqrt{13}}, \beta = \cos^{-1}\frac{3}{5}$

$\sin\alpha = \frac{2}{\sqrt{13}}, \cos\beta = \frac{3}{5}$

$\cos\alpha = \sqrt{1 - (\frac{2}{\sqrt{13}})^2} = \sqrt{1 - \frac{4}{13}} = \sqrt{\frac{9}{13}} = \frac{3}{\sqrt{13}}, \sin\beta = \sqrt{1 - (\frac{3}{5})^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$

$\tan\alpha = \frac{\sin\alpha}{\cos\alpha} = \frac{2/\sqrt{13}}{3/\sqrt{13}} = \frac{2}{3}, \tan\beta = \frac{\sin\beta}{\cos\beta} = \frac{4/5}{3/5} = \frac{4}{3}$

$\tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta} = \frac{\frac{2}{3} + \frac{4}{3}}{1 - \frac{2}{3}\cdot\frac{4}{3}} = \frac{\frac{6}{3}}{1 - \frac{8}{9}} = \frac{2}{\frac{1}{9}} = 18$

$\tan(\alpha + \beta) = 18$

$\alpha + \beta = \tan^{-1}x$

$\tan^{-1}18 = \tan^{-1}x$

$x = 18$

(4) $\cos^{-1}\frac{2}{3} + \cos^{-1}\frac{1}{4} = \sin^{-1}x$

$\alpha = \cos^{-1}\frac{2}{3}, \beta = \cos^{-1}\frac{1}{4}$

$\cos\alpha = \frac{2}{3}, \cos\beta = \frac{1}{4}$

$\sin\alpha = \sqrt{1 - (\frac{2}{3})^2} = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}, \sin\beta = \sqrt{1 - (\frac{1}{4})^2} = \sqrt{1 - \frac{1}{16}} = \sqrt{\frac{15}{16}} = \frac{\sqrt{15}}{4}$

$\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta = \frac{\sqrt{5}}{3}\cdot\frac{1}{4} + \frac{2}{3}\cdot\frac{\sqrt{15}}{4} = \frac{\sqrt{5}}{12} + \frac{2\sqrt{15}}{12} = \frac{\sqrt{5} + 2\sqrt{15}}{12}$

$\sin(\alpha + \beta) = \frac{\sqrt{5} + 2\sqrt{15}}{12}$

$\alpha + \beta = \sin^{-1}x$

$\sin^{-1}(\frac{\sqrt{5} + 2\sqrt{15}}{12}) = \sin^{-1}x$

$x = \frac{\sqrt{5} + 2\sqrt{15}}{12}$

3. 最終的な答え

(1) $x = \frac{16}{65}$

(2) $x = -\frac{2\sqrt{5}}{25}$

(3) $x = 18$

(4) $x = \frac{\sqrt{5} + 2\sqrt{15}}{12}$